Question Number 37356 by math khazana by abdo last updated on 12/Jun/18
$${let}\:\:{b}\:\in{C}\:\:{and}\:{Re}\left({b}\right)\:>\mathrm{0}\:{prove}\:{that} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{iax}} }{{x}−{ib}}{dx}\:=\mathrm{2}{i}\pi\:{e}^{−{ab}\:\:} \:\:\:{and} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{iax}} }{{x}+{ib}}\:{dx}\:=\mathrm{0} \\ $$
Commented by math khazana by abdo last updated on 12/Jun/18
$${a}>\mathrm{0} \\ $$
Commented by math khazana by abdo last updated on 13/Jun/18
$${let}\:\varphi\left({z}\right)\:=\:\frac{{e}^{{iaz}} }{{z}−{ib}}\:\:\:{so}\:{ib}\:{is}\:{a}\:{simple}\:{pole}\:{of}\:\varphi \\ $$$${with}\:{Re}\left({b}\right)>\mathrm{0}\:\Rightarrow{Im}\left({ib}\right)>\mathrm{0}\:\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{ib}\right) \\ $$$$=\mathrm{2}{i}\pi\:{e}^{{ia}\left({ib}\right)} =\:\mathrm{2}{i}\pi\:{e}^{−{ab}} \\ $$$${let}\:\psi\left({z}\right)\:=\:\frac{{e}^{{iaz}} }{{x}+{ib}}\:\:{the}\:{pole}\:{of}\:\psi\:{is}\:−{ib}\:\:{but}\:{Im}\left(−{ib}\right)<\mathrm{0} \\ $$$${so}\:\int_{−\infty} ^{+\infty} \:\:\psi\left({z}\right){dz}\:=\mathrm{0} \\ $$